I have to study the convergence of the series $$\sum_{n=1}^\infty \left(1-n\sin\left(\frac{1}{n}\right)\right).$$ I know I should study the limit $$\lim_{n \to \infty}n^\alpha\left( 1-n\sin\left(\frac{1}{n}\right)\right).$$

But I have some troubles with this limit (and so the convergence of the series). Any ideas?

  • 1
    $\begingroup$ +1 Thanks for posting this nice problem, I am going to use this problem for my students... $\endgroup$
    – imranfat
    May 12, 2015 at 14:37

2 Answers 2


HINT: $$ \sin x=x-x^3/6+\dots, $$ hence $$ \frac{\sin x}{x}-1=-x^2/6+\dots, $$ hence your series behaves as $\sum 1/n^2$.

If you are not allowed to use the Taylor expansion, you can use the ratio test, comparing with $\sum 1/n^2$.


You may observe that, using the Taylor expansion of $\sin x$ near $0$, you get, as $n \to \infty$: $$ \sin \left(\frac1n\right)=\frac1n+\mathcal{O}\left(\frac1{n^3}\right) $$ giving $$ 1-n\sin \left(\frac1n\right)=\mathcal{O}\left(\frac1{n^2}\right) $$ and the initial series $\displaystyle \sum_{n\geq1}\left(1-n\sin \left(\frac1n\right)\right)$ is convergent as is $\displaystyle \sum_{n\geq1}\frac1{n^2}.$


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